Choose the correct ascending order of the particle according to their de-broglie wavelength, when they are accelerated by same potential
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( {}^{3}_{1}\mathrm{H} \lt {}^{2}_{1}\mathrm{H} \lt {}^{1}_{1}\mathrm{H} \lt e \)
A charge \( q \) accelerated through a potential difference \( V \) gains kinetic energy \( qV \), so its momentum is \( p = \sqrt{2mqV} \) and its de Broglie wavelength is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mqV}} \)
The electron, proton \( {}^{1}_{1}\mathrm{H} \), deuteron \( {}^{2}_{1}\mathrm{H} \) and triton \( {}^{3}_{1}\mathrm{H} \) all carry a charge of magnitude \( e \), and \( V \) is the same, so
\( \displaystyle \lambda \propto \frac{1}{\sqrt{m}} \)
The masses increase in the order \( m_e \lt m_{p} \lt m_{d} \lt m_{t} \) (the nuclei have masses of about 1 u, 2 u and 3 u). The heavier particle has the shorter wavelength, so in ascending order of wavelength:
\( \displaystyle \lambda({}^{3}_{1}\mathrm{H}) \lt \lambda({}^{2}_{1}\mathrm{H}) \lt \lambda({}^{1}_{1}\mathrm{H}) \lt \lambda(e) \)
NCERT (2026-27):
Wave Nature of Matter – “… De Broglie proposed that the wave length λ associated with a particle of momentum p is given as
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{m v} \qquad (11.5) \)
where m is the mass of the particle and v its speed. … Clearly, from Eq. (11.5), λ is smaller for a heavier particle (large m) or more energetic particle (large v).”